Solutions 1 Solution : 0 1 A basis for R2 is , . Suppose that a linear transformation T : R2 → R2 is defined 2 2 0 0 1 1 0 1 as T = −3 + 2 , T = 2 + 3 . Find the matrix of 2 2 2 2 2 2 T with respect to the usual basis and the matrix with respect to the given matrix. The matrix with respect to the given basis is −3 2 2 3 The matrix with respect to the usual basis then is −1 0 1 −3 2 0 1 1 1 = 2 2 2 3 2 2 12 −1 2 Solution : 2 1 2 Let A = 3 0 −2 find det A. −1 2 2 16 3 Solution :
Suppose y1x and y2x are two solutions to the second order linear differential equation y′′x + axy′x + bxyx = 0. Let Wx denote the Wronskian of the two functions. Then show that W′x + axWx = 0 and hence Wx = Ce −Ax where A′x = ax. Thus the Wronskian either vanishes for all values of x or never. Here are two functions: 3 5 y1x = x ,y2x = x . Do there exist continuous functions ax and bx such that y1,y2 are both solutions to y′′x + axy′x + bxyx = 0 for x ∈ −1,1? Explain. ′′ ′ ′′ ′ y1 + ay 1 + by 1 = 0 and y2 + ay 2 + by 2 = 0. Then it follows that ′′ ′′ ′ ′ ′ y2 y1 − y1 y2 + ay2y1 − y1y2 = 0. Then you notice that this is of the form W + axW = 0. Multiplying both sides by eAx where A′x = ax, the result follows. No such functions exist. If so, the Wronskian could not be 0 at 0 and nonzero elsewhere. 4 Solution : 1 1 1 Let x a complex number. Find the inverse of the matrix x x − 2 x + 2 . You can x2 x − 22 x + 22 use the formula for the inverse in terms of the transpose of the cofactor matrix if you like. You also might notice that this is a Vandermonde matrix. −1 1 2 1 1 1 1 1 1 − 4 x 2 x − 4 1 2 1 1 1 1 x x − 2 x + 2 = 8 x + 4 x − 4 x − 4 8 2 2 2 1 2 1 1 1 1 x x − 2 x + 2 8 x − 4 x 4 − 4 x 8 5 Solution : −2 3 −1 4 1 −3 1 0 Let A = find det A. −1 1 4 −1 3 −2 −1 −3 −2 3 −1 4 1 −3 1 0 det = 64 −1 1 4 −1 3 −2 −1 −3 6 Solution : 2 2 −2 A basis for R3 is 1 , 2 , 2 . Suppose that a linear transformation 1 −2 2 T : R3 → R3 is defined as 2 2 2 2 2 −2 T 1 = 1 1 + 1 2 , T 2 = −2 1 + −2 2 . 1 1 −2 −2 1 2 −2 2 2 −2 T 2 = 2 1 + −1 2 + 2 2 . Find the matrix of T with respect to the 2 1 −2 2 usual basis and the matrix with respect to the given matrix. The matrix with respect to the given basis is 1 −2 2 1 0 −1 0 −2 2 The matrix with respect to the usual basis then is −1 5 1 7 2 2 −2 1 −2 2 2 2 −2 3 − 2 6 1 1 17 1 2 2 1 0 −1 1 2 2 = 3 − 2 6 5 1 11 1 −2 2 0 −2 2 1 −2 2 − 3 2 6 7 Solution : Consider the general second order differential operator D2 + aD + bI where a,b are given complex numbers. Then it is clear that this linear mapping takes functions of the form pteαt to functions of this sort. Here pt is a polynomial of degree no more than 1 and α is a complex number. Let V denote the functions of this sort. First show that a basis for V is eαt,te αt . Next find the matrix M of the linear operator with respect to this basis. Then determine values of α such that the matrix is not invertible. Finally find a solution to the differential equation D2 + aD + bI y = 0 by finding a solution to the linear equation Mx = 0. αt αt αt αt αt αt be ,te + aD te ,te + DtDte ,te = be αt + aαeαt + α2eαt,bte αt + aeαt + tαeαt + 2αeαt + tα2eαt And so we need be αt + aαeαt + α2eαt,bte αt + aeαt + tαeαt + 2αeαt + tα2eαt = eαt,te αt M where M is a 2 × 2 matrix. Thus b + aα + α2 a + 2α M = 0 b + aα + α2 Thus, we need to have b + aα + α2 = 0, Solution is : α 1 1 2 α 1 1 2 1 = − 2 a + 2 a − 4b , 2 = − 2 a − 2 a − 4b 0 a + 2α 1 Then for either value of α, M = and a solution to this is and so a 0 0 0 corresponding solution to the differential equation is y = Ce αt for either value of α. What if α α α α 1 1 = 2 = ? Then in this case = − 2 a, and so the matrix reduces to the zero matrix. Hence C anything of the form Ce αt + Dte αt will solve the differential equation because M = 0 for D any value of C,D. 8 Solution : Show that the only solutions to the second order differential equation D2 + aD + bI y = 0 are α t α t 2 αt αt C1e 1 + C2e 2 in case α1 ≠ α2 are two solutions to r + ar + b = 0 or C1e + C2te in case there is a repeated root α to this polynomial, r2 + ar + b. You only need to verify that the Wronskian of these pairs of functions is not zero at t = 0. It follows than that you can solve the equations C1 + C2 = y0,α1C1 + α2C2 = y1 in the case where α1 ≠ α2 and C1 = y0,C2 = y1. Then it follows that, using these values of Ci you obtain a ′ solution to the differential equation and the initial conditions y0 = y0,y 0 = y1. Since any solution to this differential equation has values for it and its derivative at 0, it follows that this α t α t solution must be the given linear combination of the functions. e 1 ,e 2 in case the αi are different and eαt,te αt in the case where there is only one solution to the above polynomial equation. 9 Solution : Suppose you have been able to find a solution to a second order linear differential equation ′′ ′ y + axy + bxy = 0 in the form C1y1x + C2y2x for x in some interval such that the Wronskian of the two functions does not vanish at any point. (Recall that the Wronskian either vanishes at every point or at no point from the relevant problem above. ) Suppose you want to find ′′ ′ a solution to the equation y + axy + bxy = fx. Show that there always exists a solution yp to this equation in the form y = Axy1x + Bxy2x. Then explain why every solution to ′′ ′ y + axy + bxy = f is of the form yp + C1y1x + C2y2x. This method is called the method of variation of parameters. Consider the following: ′ ′ ′ ′ ′ y = A y1 + B y2 + Ay 1 + By 2 ′ ′ Arbitrarily set A y1 + B y2 = 0. Then differentiate again to get ′′ ′ ′ ′ ′ ′′ ′′ y = A y1 + B y2 + Ay 1 + By 2 Then this will solve the equation y′′ + axy′ + bxy = f if you let ′ ′ ′ ′ A y1 + B y2 = f ′ ′ ′ ′ ′ ′ Thus we just need to solve the two equations A y1 + B y2 = 0,A y1 + B y2 = f and this can be
0 y2 y1 0 ′ ′ f y2 y1 f ′ ′ done by Cramer’s rule. Thus A = W ,B = W where W is the Wronskian of the two functions which is assumed not to vanish anywhere. Therefore, since continuous functions can be integrated, there exist Ax,Bx such that yp = Ay 1 + By 2 solves the equation ′′ ′ ′′ ′ y + axy + bxy = f. If y + axy + bxy = f is any other solution, then z = y − yp solves ′′ ′ z + axz + bxz = 0 and so there exist scalars Ci such that y = yp + C1y1x + C2y2x. 10 Solution : Find the solution to y′′ + 2y′ + y = et,y0 = 4,y′0 = 3.
1 t We look for a solution to the equation. It is easily seen to be yp = 4 e . 1 t −t −t 4 e + Ce + Dte Then we need to find the constants to solve the initial conditions. You need to solve the linear system of equations for C,D. 1 C + 4 = 4 1 D − C + 4 = 3 y 1 et 13 t 15 = 4 + 2 et + 4et 11 Solution : Find the solution to y′′ + 4y′ + 3y = sin 2t,y0 = 1,y′0 = 0. First find a solution to the equation. y′′ + 4y′ + 3y = sin 2t An easy way to do this is to look for a solution in the form y = Asin 2t + Bcos t and determine A,B. An eaiser way is to look for a solution to y′′ + 4y′ + 3y = e2it and then take the imaginary 2it part. Lets do it this way. Look for yp = Ae . Then we need to solve −1 − 8iAe 2it = e2it Then we need −1 − 8iA = 1 1 8 A = − 65 − 65 i 1 8 Then a particular solution is yp = − 65 sin 2t + − 65 cos 2t The general solution is then of the form 1 8 −t −3t − 65 sin 2t + − 65 cos 2t + Ce + De To satisfy the initial conditions, we need 8 C + D − 65 = 2 2 −C − 3D − 65 = 1 37 −t 8 41 −3t 1 y = 10 e − 65 cos 2t − 26 e − 65 sin2t 12 Solution : Find the solution to y′′ − y′ − 6y = e2t,y0 = 0,y′0 = −3. 1 32t 2t 7 2t yt = − 5e − 13e + 8e 2 20e22t 13 Solution : Find the solution to y′′ − 9y = sin 4t,y0 = 0,y′0 = 0. yt = − 1 −2e23t + 3sin4te3t + 2 75e3t 14 Solution : Find the solution to 8y + 8y′ + 4y′′ = e3t,y0 = −2,y′0 = 2.
r2 + 2r + 2 = r2 + 2r + 2 = r − −1 + ir − −1 − i ′′ ′ 3 2 and so the general solution to the equation y + 2y + y 2 = 0 is Ce −1+it + De t−1+i. It remains to find a particular solution to the equation. You can do this by the 1 3t method of variation of parameters or else simply look for one in the form 4 Ae . Thus this 1 3t particular solution is 68 e . Then the solution to the differential equation is 1 3t −t1+i t−1+i 68 e + Ce + De Now you note that the initial data is real and all the numbers involved in the equation are also real and so you should observe that the general solution is of the form 1 3t −t 68 e + e D sin t + Ccos t. You need to solve the following. 1 + C = −2 68 for C,D. 1 − 136 136C − 136D − 6 = 2 1 3t −t 137 1 y = 68 e + e − 68 cos t + − 17 sin t 15 Solution : AB Consider the block matrix, M = where A is m × m and C is k × k. Thus the matrix is 0 C m + k × m + k. Show that det M = det Adet C. If the rank of C = k so that C−1 exists, then note that, by block multiplication, AB A BC −1 I 0 = 0 C 0 I 0 C AB A BC −1 I 0 Then det = det det 0 C 0 I 0 C A 0 I 0 = det det = det Adet C. 0 I 0 C 16 Solution : n Suppose v1,⋯,vk are vectors in R . Then the volume of the parallelepiped determined by 1/2 T v1
these vectors is defined as det ⋮ v1 ⋯ vk . Verify that this makes T vk sense by showing that it gives the right answer if there is only one vector, and also gives the right answer if the vectors are othogonal. Next verify that for any list of vectors, the above determinant is nonnegative so one can at least take the square root as required. What is the answer if k > n? If
k = n, show that the expression reduces to det v1 ⋯ vk . If you only have one vector, then you are looking at vTv1/2 = 〈v,v〉1/2 which is the definition of the length of v. This is, of course the one dimensional volume of the parallelepiped spanned by v. n Next suppose you have v1,⋯,vk an orthonormal set of vectors in R . Obviously, you would want the volume of the parallelepiped to equal the product of the lengths of the vi. Is this what the T v1 × formula gives you? In this case, the k k matrix ⋮ v1 ⋯ vk reduces to the T vk T th diagonal matrix having vj vj = 〈vj,vj 〉 in the j slot on the diagonal. Therefore, the determinant is nothing more than n |v |2, and so its square root reduces to the product of the magnitudes of ∏i=1 i the vectors as required. Let V = v1 ⋯ vk . If the rank of V is less than k, then det VTV = 0 because VTV is a k × k matrix which has the same rank as V and so, if it has rank less than k, the determinant equals 0. Why is the determinant nonnegative? By the Gram Schmidt procedure, there exists an orthonormal basis for span v1,⋯,vk denoted by w1,⋯,wk and so th T T 〈vk,vj 〉 = ∑i〈vk,wi 〉〈wi,vj 〉. Thus the ij entry of V V, given by vi vj = 〈vi,vj 〉 is equal to th × ∑r〈vi,wr 〉〈wr,vj 〉 which is the ij entry of the product of k k matrix and its transpose, the th matrix having rj entry equal to 〈wr,vj 〉. Therefore, denoting this k × k matrix by A, it follows det VTV = det ATA = det A2, and so the determinant is nonnegative as claimed. If k = n then V is an n × n matrix and the determinant of the definition reduces to the square of
det v1 ⋯ vn . 17 Solution : Show that the definition of volume given in the above is the only one which is geometrically reasonable. It is clear from what was presented above that it is the only one which is geometrically reasonable if you have a single vector. Suppose this is so for p vectors in Rn, n > p. Then from the above problem, the only reasonable definition of the p + 1 dimensional volume of the parallelepiped determined by v ,⋯,v ,v is v − p 〈v,w 〉w |det Gv ,⋯,v | where G is the matrix 1 p ∑i=1 i i 1 p th whose ij entry is 〈vi,wj 〉 for wj an orthonormal basis for span v1,⋯,vp as above. Thus v − p 〈v,w 〉w is just the height of the higher dimensional parallelpiped determined by ∑i=1 i i adding in the vector v. The vector is in the direction of N or −N. You want the volume to be the height times the area of the base, as usual. Now let w v − p 〈v,w 〉w / v − p 〈v,w 〉w so that w is a unit vector perpendicular to p+1 = ∑i=1 i i ∑i=1 i i p+1 the other wi. Thus 〈v,w 〉 〈v,v〉 − p 〈v,w 〉2 / v − p 〈v,w 〉w Note p+1 = ∑i=1 i ∑i=1 i i 2 v − p 〈v,w 〉 〈v,v〉 − p 〈v,w 〉2 and so from the above, ∑i=1 i = ∑i=1 i 〈v,w 〉 v − p 〈v,w 〉w . The volume desired is therefore the absolute value of p+1 = ∑i=1 i i
〈v1,w1 〉 ⋯ 〈v1,wp 〉
〈v,wp+1 〉det ⋮ ⋮ which implies the volume is the absolute value of
〈vp,w1 〉 ⋯ 〈vp,wp 〉
0 ⋯ 0 〈v,wp+1 〉 〈v ,w 〉 ⋯ 〈v ,w 〉 0 det 1 1 1 p which is the same as the absolute value of ⋮ ⋮ ⋮
〈vp,w1 〉 ⋯ 〈vp,wp 〉 0
〈v,w1 〉 ⋯ 〈v,wp 〉〈v,wp+1 〉
〈v1,w1 〉 ⋯ 〈v1,wp 〉〈v1,wp+1 〉 det because each vi is in the span of w1,⋯,wp , and ⋮ ⋮ ⋮
〈vp,w1 〉 ⋯ 〈vp,wp 〉〈vp,wp+1 〉 so 〈vi,wp+1 〉 = 0 for each i ∈ 1,⋯,p. Expand along the last column. Thus, the p + 1 dimensional volume is det MTM1/2where M is the matrix whose determinant is given above. As 1/2 T v0 before, letting v0 = v, the desired volume is det ⋮ v0 ⋯ vp . T vp 18 Solution : Find the two dimensional volume of the parallelepiped determined by the two vectors in R4 −2,0,0,1,1,2,−2,0. 41 19 Solution : Find the three dimensional volume of the parallelepiped determined by the three vectors in R4 2,−1,0,0,0,−2,2,0,1,0,2,1. 2 2 17 20 Solution : Find the 4 dimensional volume of the parallelepiped determined by the four vectors in R3 −2,2,0,−2,2,−1,−2,0,−1,2,2,2. 0 21 Solution : n Let x1,x2,⋯,xp be vectors in R . Show that these vectors are linearly independent if and only if the parallelepiped determined by these vectors has positive p dimensional volume. If they are independent, then the matrix which has these as columns has rank p. If A is this matrix, it follows that ATA also has rank p and so its determinant is positive. Conversely, if the volume is positive, then this matrix has positive determinant, which implies A has rank p.